MHB How to decompose a fourier series

[ognik]

Hi, in a section on FS, if I were given $\sum_{n=1}^{\infty} \frac{Sin nx}{n} $ I can recognize that as the Sin component of a Fourier Series, with $b_n = \frac{1}{n} = \frac{1}{\pi} \int_{0}^{2 \pi}f(x) Sin nx \,dx$

Can I find the original f(x) from this? Differentiating both sides doesn't seem to lead anywhere?

 

[I like Serena]

Hi, in a section on FS, if I were given $\sum_{n=1}^{\infty} \frac{Sin nx}{n} $ I can recognize that as the Sin component of a Fourier Series, with $b_n = \frac{1}{n} = \frac{1}{\pi} \int_{0}^{2 \pi}f(x) Sin nx \,dx$

Can I find the original f(x) from this? Differentiating both sides doesn't seem to lead anywhere?

Hi ognik,

Typically we look it up in a list of known Fourier Series.

 

[ognik]

The exercise that prompted my question said to 'apply the summation technique of this section' (FS) 'to show that

$\sum_{n=1}^{\infty} \frac{Sin nx}{n} = \frac{1}{2}(\pi-x), 0 \lt x \le \pi; =-\frac{1}{2}(\pi+x), -\pi \le x \lt 0$

I tried differentiating the formula for $b_n$ to find the original f(x), but I don't thank that's going to work?

 

[I like Serena]

The exercise that prompted my question said to 'apply the summation technique of this section' (FS) 'to show that

$\sum_{n=1}^{\infty} \frac{Sin nx}{n} = \frac{1}{2}(\pi-x), 0 \lt x \le \pi; =-\frac{1}{2}(\pi+x), -\pi \le x \lt 0$

I tried differentiating the formula for $b_n$ to find the original f(x), but I don't thank that's going to work?

There you go. The original f(x) is already given. $f(x)=\frac 12 (\pi-x)$.
That leaves the verification that $a_n=0$ and $b_n=\frac 1n$.

 

[ognik]

The original f(x) is already given. $f(x)=\frac 12 (\pi-x)$.

I know you're right, but I can't see why! If I were told $f(x)=\frac 12 (\pi-x)$ then no problem, but I don't see the relationship in what is written...why is $ f(x)= \sum_{n=1}^{\infty}\frac{sin nx}{n} $?

 

[I like Serena]

I know you're right, but I can't see why! If I were told $f(x)=\frac 12 (\pi-x)$ then no problem, but I don't see the relationship in what is written...why is $ f(x)= \sum_{n=1}^{\infty}\frac{sin nx}{n} $?

Fourier series analysis gives us a transformation that works uniquely both ways.
From any function $f(x)$ we can calculate the coefficients $a_n$ and $b_n$, such that
$$f(x) = \frac {a_0}{2} + \sum a_n \cos nx + \sum b_n \sin nx$$
The other way around, those $a_n$ and $b_n$ uniquely determine $f(x)$.
That is:
$$f(x) \leftrightarrow (a_n,b_n)$$

Since it is already given that $ f(x)= \sum_{n=1}^{\infty}\frac 1 n{\sin nx} $, we can conclude that:
$$a_n = \frac 1\pi \int f(x) \cos nx \,dx = 0$$
$$b_n = \frac 1\pi \int f(x) \sin nx \,dx = \frac 1 n$$
The other way around, if those $a_n$ and $b_n$ are given, we know that the series sums up to $f(x)$.

 

[ognik]

Thats where I got to, wrote down the formulae for $a_n$ and $b_n$ with f(x) within them. However, unless I recognized what f(x) must be from previous experience, or looking it up in a table, I couldn't find f(x) from $b_n) analytically?

Are you saying that if I sketched the Fourier series, it would be easy to recognize the original function?

 

[I like Serena]

Thats where I got to, wrote down the formulae for $a_n$ and $b_n$ with f(x) within them. However, unless I recognized what f(x) must be from previous experience, or looking it up in a table, I couldn't find f(x) from $b_n) analytically?

Are you saying that if I sketched the Fourier series, it would be easy to recognize the original function?

We can do it analytically by calculating the summation.
And yes, by graphing the sum of the first couple of terms, we can also recognize the function.

 

[ognik]

We can do it analytically by calculating the summation.

Thanks ILS, but I don't understand how to calculate $ \sum_{n=1}^{\infty} \frac{Sin nx}{n} $? I can write out the first few terms but they don't point back to $\frac{1}{2}(\pi - x) $ etc. ?

 

[I like Serena]

Thanks ILS, but I don't understand how to calculate $ \sum_{n=1}^{\infty} \frac{Sin nx}{n} $? I can write out the first few terms but they don't point back to $\frac{1}{2}(\pi - x) $ etc. ?

Well, I guess it is not an easy sum to find.
There are probably special techniques of doing it, but to be honest, I wouldn't know off the bat how to do it either.
... except with the trick we just used by considering it as a Fourier series, which is a valid way to find the result analytically and it's easy enough to expand $\frac{1}{2}(\pi - x) $ as a Fourier series.